POJ 3259 Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 40699 Accepted: 14914

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N,M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps toF (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S,E, T) that describe, respectively: A one way path from S toE that also moves the traveler back T seconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

带负圈的最短路径问题，用Bellman_Ford算法或者SPFA算法都能过

#include <stdio.h>#include <string.h>int N, M, W;struct node{int u;int v;int w;} map[5500];int dist[505];bool Bellman_Ford(){int i, j;for(i = 1; i < N; i++)for(j = 1; j <= 2*M+W; j++)if(dist[map[j].u] + map[j].w < dist[map[j].v])dist[map[j].v] = dist[map[j].u] + map[j].w;for(i = 1; i <= 2*M+W; i++)if(dist[map[i].u] + map[i].w < dist[map[i].v])return true;return false;}int main(){int F, i, j, S, E, T;scanf("%d", &F);while(F--){memset(dist, 0x3f, sizeof(dist));scanf("%d%d%d", &N, &M, &W);for(i = 1; i <= M; i++){scanf("%d%d%d", &S, &E, &T);map[2*i-1].u = S;map[2*i-1].v = E;map[2*i-1].w = T;map[2*i].u = E;map[2*i].v = S;map[2*i].w = T;}for(i = 1; i <= W; i++){scanf("%d%d%d", &S, &E, &T);map[2*M+i].u = S;map[2*M+i].v = E;map[2*M+i].w = -T;}dist[1] = 0;if(Bellman_Ford())printf("YES\n");elseprintf("NO\n");}return 0;}